Operations on the Set of Real Numbers

Lessons

Senior 4 · Mathematics

Operations on the Set of Real Numbers

5 lessons

Lesson 3 of 5

Operations on radicals and rationalising the denominator

25 min

Exact answers that can be added

The triangle below has its corners on grid points. Pythagoras gives its sides exactly: $AB = \sqrt{4^2 + 2^2} = \sqrt{20}$ , $BC = \sqrt{2^2 + 2^2} = \sqrt{8}$ and $CA = \sqrt{2^2 + 4^2} = \sqrt{20}$ . An exam asks for the perimeter "in simplest surd form". To answer, you must know which radicals can be added, how to multiply them, and how to clear them out of a denominator. Those three skills are this lesson.

Definition

Like radicals: radicals with the same index and the same radicand, such as $3\sqrt[3]{2}$ and $-5\sqrt[3]{2}$ . Only like radicals can be added or subtracted, by adding their coefficients: $3\sqrt[3]{2} - 5\sqrt[3]{2} = -2\sqrt[3]{2}$ .

Conjugate: the conjugate of $a + b\sqrt{c}$ is $a - b\sqrt{c}$ , and the conjugate of $\sqrt{p} + \sqrt{q}$ is $\sqrt{p} - \sqrt{q}$ . Their product has no radical:

(a + b\sqrt{c})(a - b\sqrt{c}) = a^2 - b^2 c .

Worked example 1: adding and subtracting

(a) Find the perimeter of triangle $A B C$ in simplest form. (b) Simplify $3\sqrt{12} - \sqrt{75} + \sqrt{27}$ . (c) Simplify $\sqrt[3]{16} + \sqrt[3]{54}$ .

Solution Simplify each radical first; only then can you see which are alike.

$\sqrt{20} = 2\sqrt{5}$ and $\sqrt{8} = 2\sqrt{2}$ , so the perimeter is $2\sqrt{5} + 2\sqrt{2} + 2\sqrt{5} = 4\sqrt{5} + 2\sqrt{2} \approx 11.77$ units. The terms $4\sqrt{5}$ and $2\sqrt{2}$ are not alike, so the answer stops there.
$3\sqrt{12} = 3 \times 2\sqrt{3} = 6\sqrt{3}$ , $\sqrt{75} = 5\sqrt{3}$ , $\sqrt{27} = 3\sqrt{3}$ . Total: $(6 - 5 + 3)\sqrt{3} = 4\sqrt{3}$ .
$\sqrt[3]{16} = \sqrt[3]{8 \times 2} = 2\sqrt[3]{2}$ and $\sqrt[3]{54} = \sqrt[3]{27 \times 2} = 3\sqrt[3]{2}$ . Total: $5\sqrt[3]{2}$ .

Warning

There is no sum rule for roots. $\sqrt{a + b} \ne \sqrt{a} + \sqrt{b}$ in general: $\sqrt{9 + 16} = \sqrt{25} = 5$ , but $\sqrt{9} + \sqrt{16} = 7$ . The product and quotient rules are the only ones that let you split a radical.

Multiplying and dividing radicals

Same index: use the product rule, $\sqrt[n]{a}\,\sqrt[n]{b} = \sqrt[n]{ab}$ , and expand brackets as in algebra.
Different indices: bring both to the lowest common index first. For example, $\sqrt{2}\,\sqrt[3]{3} = \sqrt[6]{2^3}\,\sqrt[6]{3^2} = \sqrt[6]{8 \times 9} = \sqrt[6]{72}$ .

Expanding a product of two binomials follows FOIL, with $\sqrt{a}\,\sqrt{a} = a$ for $a \ge 0$ :

(2\sqrt{3} - \sqrt{5})(\sqrt{3} + 2\sqrt{5}) = 2 \cdot 3 + 4\sqrt{15} - \sqrt{15} - 2 \cdot 5 = -4 + 3\sqrt{15} .

Definition

Rationalising the denominator: rewriting a fraction so that its denominator contains no radical, by multiplying the numerator and the denominator by the same suitable factor (the rationalising factor). The value of the fraction does not change, because you multiply by 1.

Example: $\frac{1}{\sqrt{2}} = \frac{1 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{\sqrt{2}}{2} \approx 0.7071$ . Halving 1.4142 by hand is far easier than dividing 1 by 1.4142, which is why exact answers are written this way.

Choosing the rationalising factor

Denominator	Multiply top and bottom by	Because
$\sqrt{b}$	$\sqrt{b}$	$\sqrt{b}\,\sqrt{b} = b$
$\sqrt[n]{b^k}$ , $0 \lt k \lt n$	$\sqrt[n]{b^{n-k}}$	$\sqrt[n]{b^k}\,\sqrt[n]{b^{n-k}} = \sqrt[n]{b^n} = b$ ( $b \gt 0$ )
$a + \sqrt{b}$ or $\sqrt{a} + \sqrt{b}$	the conjugate	$(x + y) (x - y) = x^{2} - y^{2}$
$\sqrt[3]{a} - \sqrt[3]{b}$	$\sqrt[3]{a^2} + \sqrt[3]{ab} + \sqrt[3]{b^2}$	$(x - y) (x^{2} + x y + y^{2}) = x^{3} - y^{3}$
$\sqrt[3]{a} + \sqrt[3]{b}$	$\sqrt[3]{a^2} - \sqrt[3]{ab} + \sqrt[3]{b^2}$	$(x + y) (x^{2} - x y + y^{2}) = x^{3} + y^{3}$

Worked example 2: one-term denominators

Rationalise (a) $\frac{10}{\sqrt{5}}$ (b) $\frac{6}{\sqrt[3]{4}}$ .

Solution

$\frac{10}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = \frac{10\sqrt{5}}{5} = 2\sqrt{5}$ .
$\sqrt[3]{4} = \sqrt[3]{2^2}$ , so the factor is $\sqrt[3]{2}$ : $\frac{6\sqrt[3]{2}}{\sqrt[3]{2^3}} = \frac{6\sqrt[3]{2}}{2} = 3\sqrt[3]{2}$ . (Multiplying by $\sqrt[3]{4}$ would give $\sqrt[3]{16}$ in the denominator, which is still irrational.)

Worked example 3: conjugates

Rationalise (a) $\frac{4}{3 - \sqrt{5}}$ (b) $\frac{\sqrt{7} + \sqrt{3}}{\sqrt{7} - \sqrt{3}}$ .

Solution

$\frac{4}{3 - \sqrt{5}} \times \frac{3 + \sqrt{5}}{3 + \sqrt{5}} = \frac{4(3 + \sqrt{5})}{9 - 5} = \frac{4(3 + \sqrt{5})}{4} = 3 + \sqrt{5}$ .
Multiply by $\frac{\sqrt{7} + \sqrt{3}}{\sqrt{7} + \sqrt{3}}$ . The denominator becomes $7 - 3 = 4$ and the numerator $(\sqrt{7} + \sqrt{3})^2 = 7 + 2\sqrt{21} + 3 = 10 + 2\sqrt{21}$ . So the fraction is $\frac{10 + 2\sqrt{21}}{4} = \frac{5 + \sqrt{21}}{2}$ .

Worked example 4: a cube-root denominator

Rationalise $\frac{1}{\sqrt[3]{2} - 1}$ .

Solution With $x = \sqrt[3]{2}$ and $y = 1$ , the identity $(x - y) (x^{2} + x y + y^{2}) = x^{3} - y^{3}$ says the factor is $\sqrt[3]{4} + \sqrt[3]{2} + 1$ :

\frac{1}{\sqrt[3]{2} - 1} \times \frac{\sqrt[3]{4} + \sqrt[3]{2} + 1}{\sqrt[3]{4} + \sqrt[3]{2} + 1} = \frac{\sqrt[3]{4} + \sqrt[3]{2} + 1}{2 - 1} = \sqrt[3]{4} + \sqrt[3]{2} + 1 .

Check: $\frac{1}{1.2599 - 1} \approx 3.847$ and $1.5874 + 1.2599 + 1 = 3.847$ . ✓

Tip

Before multiplying out, ask: "What times this denominator gives a whole number?" For $\sqrt{b}$ it is $\sqrt{b}$ ; for a sum or difference of square roots it is the conjugate; for cube roots, use the sum or difference of cubes. In Unit 7 you will use the same trick the other way round, rationalising a numerator such as $\sqrt{x + 1} - 1$ to find a limit.

The Organic Chemistry Tutor: Rationalize The Denominator. Watch for how the factor is chosen for one-term denominators, for cube roots, and for binomials with the conjugate.

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Practice

Give every answer in simplest form.

Simplify: (a) $2\sqrt{18} + \sqrt{50} - \sqrt{32}$ (b) $\sqrt[3]{81} - \sqrt[3]{24}$ .
Expand and simplify: (a) $(3 + \sqrt{2})(5 - \sqrt{2})$ (b) $(\sqrt{6} - \sqrt{3})^2$ .
Write $\sqrt{3}\,\sqrt[3]{2}$ as a single radical.
Rationalise: (a) $\frac{14}{\sqrt{7}}$ (b) $\frac{3}{\sqrt[3]{9}}$ (c) $\frac{1}{\sqrt{5} - 2}$ .
Rationalise $\frac{2\sqrt{3} + 1}{2\sqrt{3} - 1}$ .
Rationalise $\frac{1}{\sqrt[3]{3} + 1}$ .
Show that $(2 + \sqrt{3})^2 = 7 + 4\sqrt{3}$ . Hence write $\frac{1}{\sqrt{7 + 4\sqrt{3}}}$ in the form $a + b\sqrt{3}$ .
Triangle $P Q R$ has $P (0, 0)$ , $Q (6, 3)$ and $R (2, 4)$ . Find its perimeter in simplest surd form.

Show answersHide answers

(a) $6\sqrt{2} + 5\sqrt{2} - 4\sqrt{2} = 7\sqrt{2}$ (b) $3\sqrt[3]{3} - 2\sqrt[3]{3} = \sqrt[3]{3}$ .
(a) $15 - 3\sqrt{2} + 5\sqrt{2} - 2 = 13 + 2\sqrt{2}$ (b) $6 - 2\sqrt{18} + 3 = 9 - 6\sqrt{2}$ .
$\sqrt[6]{3^3}\,\sqrt[6]{2^2} = \sqrt[6]{108}$ .
(a) $\frac{14\sqrt{7}}{7} = 2\sqrt{7}$ (b) $\frac{3\sqrt[3]{3}}{\sqrt[3]{27}} = \sqrt[3]{3}$ (c) $\frac{\sqrt{5} + 2}{5 - 4} = \sqrt{5} + 2$ .
Multiply by $2\sqrt{3} + 1$ : $\frac{(2\sqrt{3} + 1)^2}{12 - 1} = \frac{13 + 4\sqrt{3}}{11}$ .
Multiply by $\sqrt[3]{9} - \sqrt[3]{3} + 1$ : the denominator is $3 + 1 = 4$ , so the answer is $\frac{\sqrt[3]{9} - \sqrt[3]{3} + 1}{4}$ .
$(2 + \sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}$ , and $2 + \sqrt{3} \gt 0$ , so $\sqrt{7 + 4\sqrt{3}} = 2 + \sqrt{3}$ . Then $\frac{1}{2 + \sqrt{3}} = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}$ .
$PQ = \sqrt{45} = 3\sqrt{5}$ , $QR = \sqrt{16 + 1} = \sqrt{17}$ , $RP = \sqrt{20} = 2\sqrt{5}$ . Perimeter $5\sqrt{5} + \sqrt{17} \approx 15.30$ .

Quick check

Simplify √12 + √27.

Key points

Simplify every radical first; only like radicals (same index, same radicand) can be added.
Multiply radicals with different indices by bringing them to a common index.
Rationalise with $\sqrt[n]{b^{n-k}}$ , with the conjugate, or with the sum or difference of cubes.
Check a rationalised answer with a calculator: its value must not change.

Sources and credits

Written by Ganzaa: the explanations, the diagram, the worked examples, the practice items and the questions.
Video: Rationalize The Denominator, The Organic Chemistry Tutor, embedded from YouTube.

1Exact answers that can be added

2Multiplying and dividing radicals

3Choosing the rationalising factor

Exact answers that can be added

Multiplying and dividing radicals

Choosing the rationalising factor